Q.If $\sin\theta + \cos\theta = 1$, then find the general value of $\theta$.

The key idea is to transform the given equation into a single trigonometric ratio using the auxiliary angle method.

Given the equation $\sin\theta + \cos\theta = 1$.

1. Divide both sides by $\sqrt{1^2+1^2} = \sqrt{2}$: $\frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta = \frac{1}{\sqrt{2}}$
2. Recognize $\frac{1}{\sqrt{2}}$ as $\cos\frac{\pi}{4}$ and $\sin\frac{\pi}{4}$. Substitute these values: $\cos\frac{\pi}{4}\sin\theta + \sin\frac{\pi}{4}\cos\theta = \sin\frac{\pi}{4}$
3. Apply the trigonometric identity $\sin A \cos B + \cos A \sin B = \sin(A+B)$: $\sin\left(\theta + \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right)$
4. Use the general solution for $\sin x = \sin \alpha$, which is $x = n\pi + (-1)^n \alpha$, where $n \in \mathbb{Z}$: $\theta + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$

Consider two cases for $n$:

• If $n$ is an even integer (let $n=2k$ for $k \in \mathbb{Z}$): …

The problem asks for the general value of $\theta$ that satisfies the equation $\sin\theta + \cos\theta = 1$. The most effective way to solve equations involving a sum of sine and cosine terms is to transform the expression $a\sin\theta + b\cos\theta$ into a single trigonometric function of the form $R\sin(\theta + \alpha)$ or $R\cos(\theta - \alpha)$. This transformation simplifies the equation into a basic trigonometric form, which can then be solved using standard general solution formulas.

The intuition behind this transformation is that any point $(a, b)$ in the Cartesian plane can be represented in polar coordinates $(R, \alpha)$, where $R = \sqrt{a^2+b^2}$ is the distance from the origin and $\alpha$ is the angle with the positive x-axis. When we factor out $R$ from $a\sin\theta + b\cos\theta$, we get $R\left(\frac{a}{R}\sin\theta + \frac{b}{R}\cos\theta\right)$. We can then identify $\frac{a}{R}$ as $\cos\alpha$ and $\frac{b}{R}$ as $\sin\alpha$ (or vice-versa), allowing us to use the angle sum/difference identities for sine or cosine.

1. Transform the expression $\sin\theta + \cos\theta$:

   We have the expression $1\cdot\sin\theta + 1\cdot\cos\theta$. This is of the form $a\sin\theta + b\cos\theta$ where $a=1$ and $b=1$.

   We calculate $R = \sqrt{a^2+b^2} = \sqrt{1^2+1^2} = \sqrt{2}$.

   Now, we can write:

   $\sin\theta + \cos\theta = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta\right)$

   We need to find an angle $\alpha$ such that $\cos\alpha = \frac{1}{\sqrt{2}}$ and $\sin\alpha = \frac{1}{\sqrt{2}}$. The principal value for such an angle is $\alpha = \frac{\pi}{4}$.

   Substituting these values, we get:

   $\sqrt{2}\left(\cos\frac{\pi}{4}\sin\theta + \sin\frac{\pi}{4}\cos\theta\right)$

   The angle sum identity for sine is $\sin(A+B) = \sin A \cos B + \cos A \sin B$.

   Using this identity, the expression becomes:

   $\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)$

   So, the original equation $\sin\theta + \cos\theta = 1$ transforms into:

   $\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) = 1$

2. Solve the basic trigonometric equation:

   Divide by $\sqrt{2}$ to isolate the sine term:

   $\sin\left(\theta + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

   We know that $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

   So, the equation is $\sin\left(\theta + \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right)$.

   The general solution for $\sin x = \sin y$ is given by $x = n\pi + (-1)^n y$, where $n \in \mathbb{Z}$ (the set of all integers).

   Applying this formula, with $x = \theta + \frac{\pi}{4}$ and $y = \frac{\pi}{4}$:

   $\theta + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$

3. Express the general value of $\theta$: …